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Answer to #1. L source-receptor distance i hydraulic gradient K hydraulic conductivity n effective soil porosity BD soil bulk density foc fraction organic carbon Koc organic partition coefficient Porosity and conductivity would likely be positively related

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L source-receptor distance

K hydraulic conductivity

n effective soil porosity

BD soil bulk density

foc fraction organic carbon

Koc organic partition coefficient

Porosity and conductivity would likely be positively related

Porosity and density would likely be inversely related

Thus, conductivity and density should be inversely related too

Fraction organic carbon and the partition coefficient would likely be positively related

L = [80,120] m // source-receptor distance

i = [0.0003,0.0008] m per m // hydraulic gradient

K = [300,3000] m per yr // hydraulic conductivity

n = [0.2,0.35] // effective soil porosity

BD = [1500,1750] kg per m3 // soil bulk density

foc = [0.0001,0.005] // fraction organic carbon

Koc = [5 ,20] m3 per kg // organic partition coefficient

Kd = foc * Koc

R = 1 + BD * Kd / n

V = K * i / (n * R)

T = L/V

T

[ 20.952380952, 408800] yr

func TT() return L/(K * i / (\$1 * (1 + BD * foc * Koc / \$1)))

b = left(n)

e = TT(b)

w = width(n)/100

for k=1 to 100 do begin

s = [ (k-1), k] * w + b

e = env(e, TT(s))

end

e

[ 31.480562448, 235352] yr

Iwater = [1.5, 2.5] liters per day // water intake

Ifish = [0, 8] g per day // dietary ingestion of fish tissue

B = [0.9, 2.1] liters per g // bioaccumulation factor

W = [60, 90] kg // receptor biomass

D = [0, 6] mg per kg per day // tolerable dose

// forward equation

// D = (Iwater * C) / W + (Ifish * B * C) / W

// condense repeated parameters

// D = C * (Iwater + Ifish * B) / W

// solve for C by factoring the quotient (Iwater+Ifish*B)/W out of D

C = factor((Iwater + Ifish * B) / W, D)

C

[ 0, 18.652849741] mg liters1

// put C back into original forward equation (and ensure correct units)

d = (Iwater * C) / W + (Ifish * C * B) / W + 0 mg per kg per day

d

[ 0, 6] mg kg1 day1

3.5105

[3.499105, 3.504105]

[3.50105, 1.35104]

[3105, 1.4104]

[2.9105, 4.1105]

[ 2.5105, 1.905 104]

Vesely et al. (all independent)

Mixed dependencies

Frank to Fréchet

All Fréchet

Mixed dependence with intervals

All Fréchet

105

104

103

Probability of tank rupturing under pumping

plane =random

ship=random

raft =random

pet =random

zoo=random

pregnant =random

malefemale=random

repro=random

female =random

parthenog=random

dogs =random

cats =random

pigs=random

eggs =random

rodents=random

hides =random

quick =random

disperse=random

(plane ||| ship ||| raft ||| pet ||| zoo) |&| (pregnant ||| (malefemale |&| repro) ||| (female |&| parthenog)) |&| (dogs |&| cats |&| pigs) |&| (eggs ||| rodents) |&| (hides ||| quick ||| disperse)

• Use the events from the slide “Snakes on a plane”

• Write as Boolean expression

• Represent ANDs as conjunctions &

• Represent ORs as disjunctions V

(plane V ship V raft V pet V zoo) & (pregnant V (malefemale & repro) V (female & parthenog)) & (dogs & cats & pigs) & (eggs V rodents) & (hides V quick V disperse)

• Specify the temporal context, e.g., a year

• Estimate marginal event probabilities

• Maybe as submodels, e.g., plane = flights * stowaway per flight

• Decide which events are independent

Assuming a = 0.29, b = 0.22, and r = 1.0, yields

a * b + r * sqrt(a*(1-a)*b*(1-b)) Pearson AND

0.2517692528

[ max(0, a+b–1), min(a, b) ] Fréchet AND

[ 0, 0.22]

The probability assuming Pearson correlation r =1 is larger than the largest possible probability. This happens because the Pearson correlation for two events with these marginals cannot be as large as one (so we are making a false assumption which leads to the error). The extreme Pearson correlations for them are

(-a*b)/sqrt(a*b*(1-a)*(1-b)) // when they don’t overlap

-0.33941721344

(b-a*b)/sqrt(a*b*(1-a)*(1-b)) // when b’s inside a

0.83098697084

We can check this with this calculation

r = 0.83098697084

a * b + r * sqrt(a*(1-a)*b*(1-b))

0.22